Bash 使用 subshel​​l 和 substring 进行错误替换

一个人为的例子......给出

FOO="/foo/bar/baz"

这有效（在 bash 中）

BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1}       # result is BAZ="b"

这不

BAZ=${$(basename $FOO):0:1} # result is bad substitution

我的问题是哪个规则导致这个[子shell替换]评估不正确？在 1 跳中执行此操作的正确方法（如果有）是什么？

首先，请注意，当你说这句话时：

BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1}       # result is BAZ="b"

构造中的第一位BAZ is BAR而不是value您想要获取其第一个字符。因此，即使 bash 允许变量名包含任意字符，第二个表达式中的结果也不会是您想要的。

然而，至于阻止这种情况的规则，请允许我引用 bash 手册页：

DEFINITIONS
       The following definitions are used throughout the rest  of  this  docu‐
       ment.
       blank  A space or tab.
       word   A  sequence  of  characters  considered  as a single unit by the
              shell.  Also known as a token.
       name   A word consisting only of  alphanumeric  characters  and  under‐
              scores,  and beginning with an alphabetic character or an under‐
              score.  Also referred to as an identifier.

稍后：

PARAMETERS
       A parameter is an entity that stores values.  It can be a name, a  num‐
       ber, or one of the special characters listed below under Special Param‐
       eters.  A variable is a parameter denoted by a name.  A variable has  a
       value  and  zero or more attributes.  Attributes are assigned using the
       declare builtin command (see declare below in SHELL BUILTIN COMMANDS).

稍后当它定义您要询问的语法时：

   ${parameter:offset:length}
          Substring Expansion.  Expands to  up  to  length  characters  of
          parameter  starting  at  the  character specified by offset.

因此，手册页中阐明的规则表明${foo:x:y}构造必须有一个参数作为第一部分，并且参数只能是名称、数字或少数特殊参数字符之一。$(basename $FOO)不是参数允许的可能性之一。

至于在一个作业中执行此操作的方法，请使用其他响应中提到的其他命令的管道。